On 12 Juni, 06:50, JT <jonas.thornv...@gmail.com> wrote:> On 11 Juni, 22:23, rich...@cogsci.ed.ac.uk (Richard Tobin) wrote:>> > In article <22cede7a-ba1d-4e00-9829-8eef91562...@g9g2000vbl.googlegroups.com>,>> > JT <jonas.thornv...@gmail.com> wrote:> > >Well 5!*3!*7 = 5040 and 7!=5040, tomorrow i give it another try.>> > Obviously, since 7! = 5! * 6 * 7, and 3! = 6. There is nothing> > interesting in this - one of the numbers greater than 7 and> > less-than-or-equal to 9 just happens to be a factorial. It's no more> > significant than that 5040! = 5039! * 7!.>> > -- Richard>> The reason i looking for a pattern is i will try implement a> permutation algorithm for elements 1-n.>> I have seen there is Steinhaus, Johnsson, Trotter algorithm that> someone directed me to, my will probably be something similar working> topdown from biggest to smallest using pairs.> When looking at the pairs for the 5! permutation, it seem that each> unique base pair 54,45,32 etc can produce 3! of subpairs.>> So for each of the 20 base pair> 54,53,52,51,45,43,42,35,34,32,31,25,24,23,21,15,14,13,12,11 there is> just 3! digits to play with, i was thinking this must be a working> approach even for 7! or any number of permutations.>> First some combinatorial finding out how many ways to select 2 among> 7, and from there is must follow that for each of those basepairs> there can be 5! of combinatorial subpairs.> I think that is the correct approach.>> 1. Find out how many ways 2 first digits can be combined within the> the group of n digits and write out.> 2. n-2 digit to calculate possible number of subpairs for each> basepair, create write out subpair.> 3. Repeat step 1 until n=0.>> 54 32 1> 54 31 2> 54 23 1> 54 21 3> 54 13 2> 54 12 3